Question

The condition of parallelism that is put to use while drawing a line parallel to a given line as shown in the construction is
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Answer 1

Step-by-step explanation:

Condition of Parallelism of Lines

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We will learn how to find the condition of parallelism of lines.

If two lines of slopes m1 and m2 are parallel, then the angle θ between them is of 90°.

Therefore, tan θ = tan 0° = 0

⇒ m2−m11+m1m2 = 0, [Using tan θ = ± m2−m11+m1m2]

⇒ m2−m1 = 0

⇒ m2 = m1

⇒ m1 = m2

Thus when two lines are parallel, their slopes are equal.

Let, the equations of the straight lines AB and CD are y = m1x+ c1 and y = m2x + c2 respectively.

If the straight lines AB and CD be parallel, then we shall have m1 = m2.

That is the slope of line y = m1 x+ c1 = the slope of the line y = m2x + c2

Conversely, if m1 = m2 then the lines y = m1 x+ c1 and y = m2x + c2 make the same angle with the positive direction of x-axis and hence, the lines are parallel.

Solved examples to find the condition of parallelism of two given straight lines:

1. What is the value of k so that the line through (3, k) and (2, 7) is parallel to the line through (-1, 4) and (0, 6)?

Solution:

Let A(3, k), B(2, 7), C(-1, 4)and D(0, 6) be the given points. Then,

m1 = slope of the line AB = 7−k2−3 = 7−k−1 = k -7

m2 = slope of the line CD = 6−40−(−1) = 21 = 2

Since, Ab and CD are parallel, therefore = slope of the line AB = slope of the line CD i.e., m1 = m2.

Thus,

k - 7 = 2

Adding 7 on both sides we get,

K - 7 + 7 = 2 + 7

K = 9

Therefore, the value of k = 9.

2. A quadrilateral has the vertices at the points (-4, 2), (2, 6), (8, 5) and (9, -7). Show that the mid-points of the sides of this quadrilateral are the vertices of a parallelogram.

Solution:

Let A(-4, 2), B(2, 6), C(8, 5) and D(9, -7) be the vertices of the given quadrilateral. Let P,Q, R and S be the mid-points of AB, BC, CD and DA respectively. Then the coordinates of P, Q, R and S are P(-1, 4), Q (5, 11/2), R(17/2, -1) and S(5/2, -5/2).

In order to prove that PQRS is a parallelogram, it is sufficient to show that PQ is parallel to RS and PQ =RS.

We have, m1 = Slope of the side PQ = 112−45−(−1)= ¼

m2 = Slope of the side RS = −52+152−172 = ¼

Clearly, m1 = m2. This shows that PQ is parallel to RS.

Now, PQ = (5+1)2+(112−4)2−−−−−−−−−−−−−−−−√ = √1532

RS = (52−172)2+(−52+1)2−−−−−−−−−−−−−−−−−−√ = √1532

Therefore, PQ = RS

Thus PQ ∥ RS and PQ = RS.

Hence, PQRS is a parallelogram.